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Determine the cumulative distribution function of a binomial random variable with \(n=3\) and \(p=1 / 2\).

Short Answer

Expert verified
The CDF is a piecewise function with probabilities 0, \(\frac{1}{8}\), \(\frac{1}{2}\), \(\frac{7}{8}\), and 1.

Step by step solution

01

Understanding the Problem

We need to find the cumulative distribution function (CDF) for a binomial random variable. Here, the number of trials \( n \) is 3, and the probability of success \( p \) is \( \frac{1}{2} \). The CDF gives the probability that the binomial random variable \( X \) is less than or equal to a certain value \( x \).
02

Formula for the Probability of Each Outcome

For a binomial random variable, the probability of \( X = k \), where \( 0 \leq k \leq n \), is given by \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \). Thus, for \( n=3 \) and \( p=\frac{1}{2} \), we calculate: \( P(X = 0) \), \( P(X = 1) \), \( P(X = 2) \), and \( P(X = 3) \).
03

Calculate Individual Probabilities

Calculate the probabilities for each value of \( k \).- \( P(X = 0) = \binom{3}{0} \left(\frac{1}{2}\right)^0 \left(\frac{1}{2}\right)^3 = \frac{1}{8} \).- \( P(X = 1) = \binom{3}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^2 = \frac{3}{8} \).- \( P(X = 2) = \binom{3}{2} \left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^1 = \frac{3}{8} \).- \( P(X = 3) = \binom{3}{3} \left(\frac{1}{2}\right)^3 \left(\frac{1}{2}\right)^0 = \frac{1}{8} \).
04

Calculate the Cumulative Distribution Function (CDF)

The CDF, \( F(x) \), is calculated as \( F(x) = P(X \leq x) \). Thus:- \( F(0) = P(X \leq 0) = P(X = 0) = \frac{1}{8} \).- \( F(1) = P(X \leq 1) = P(X = 0) + P(X = 1) = \frac{1}{8} + \frac{3}{8} = \frac{4}{8} = \frac{1}{2} \).- \( F(2) = P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = \frac{1}{2} + \frac{3}{8} = \frac{7}{8} \).- \( F(3) = P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1 \).
05

Final Result

The CDF of the binomial random variable with \( n = 3 \) and \( p = \frac{1}{2} \) is:\[ F(x) = \begin{cases} 0, & \text{if } x < 0, \ \frac{1}{8}, & \text{if } 0 \leq x < 1, \ \frac{1}{2}, & \text{if } 1 \leq x < 2, \ \frac{7}{8}, & \text{if } 2 \leq x < 3, \ 1, & \text{if } x \geq 3.\end{cases} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Distribution
The binomial distribution is one of the simplest and quite frequently used discrete probability distributions in statistics. It models the number of successes in a fixed number of trials, with only two possible outcomes in each trial, often termed as "success" or "failure."

The key parameters that define a binomial distribution are:
  • The number of trials, denoted by \( n \).
  • The probability of success in each trial, denoted by \( p \).
In our exercise, \( n = 3 \) and \( p = \frac{1}{2} \). This means we are looking at three trials, each with a 50% chance of success.

When using a binomial distribution, we can calculate the probability of achieving a certain number of successes using the binomial probability formula:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]where \( \binom{n}{k} \) is the binomial coefficient, representing the number of ways to choose \( k \) successes from \( n \) trials.
Probability
At its core, probability is a measure of the likelihood of a particular event occurring. It is expressed as a value between 0 and 1, with 0 indicating that an event is impossible and 1 indicating that it is certain. In the context of binomial distribution, probability helps in figuring out the chance of achieving a particular outcome from a series of events.

In a binomial setup, each trial is independent, and for each trial, the probability of success stays consistent. This consistent probability is a central aspect of binomial distribution calculations. For instance, in a coin toss scenario with \( p = \frac{1}{2} \), each flip of the coin is equally likely to result in heads or tails. This same probability principle is at work in our exercise where \( p \), probability of success, is \( \frac{1}{2} \), and plays a crucial role in calculating the cumulative distribution function (CDF).

By understanding the probability formula of each outcome, it becomes efficient to compute the likelihood of different numbers of successful outcomes.
Random Variables
Random variables are fundamental in probability theory and statistics. They are used to describe outcomes of random phenomena, and can be either discrete or continuous. In a binomial distribution like ours, the random variable, often denoted as \( X \), is discrete since it can take only a finite number of values.

For our exercise, the random variable \( X \) represents the number of successes in 3 trials. \( X \) can take on values 0, 1, 2, or 3, corresponding to the different numbers of successful outcomes in the three trials. Each outcome has a specific probability, which can be computed using the binomial formula.

Random variables help us bridge the theoretical probabilities with actual outcomes, allowing us to analyze and interpret probabilities in a structured manner. Importantly, each calculated probability for \( X = k \) represents the likelihood of obtaining exactly \( k \) successes.
Discrete Probability Distributions
A discrete probability distribution describes the likelihood of each of the possible outcomes of a discrete random variable. It assigns a probability to each possible value.
  • The sum of all probabilities in a discrete distribution equals 1, following the rule that the total probability of all possible outcomes must be complete.
  • The binomial distribution is one particular example of a discrete probability distribution.
In our scenario, we have a clear discrete set of outcomes resulting from a binomial experiment.

Each probability calculated for \( X = 0, 1, 2, \) or 3 in the original exercise aligns with the concept of a discrete probability distribution. Together these probabilities form the probability mass function (PMF) of our binomial variable.

The cumulative distribution function (CDF) extends from the PMF by showing cumulative probabilities. It shows the probability that the random variable is less than or equal to a particular value, offering a useful summary of the distribution dynamics in a cumulative manner.

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Most popular questions from this chapter

Each main bearing cap in an engine contains four bolts. The bolts are selected at random, without replacement, from a parts bin that contains 30 bolts from one supplier and 70 bolts from another. (a) What is the probability that a main bearing cap contains all bolts from the same supplier? (b) What is the probability that exactly three bolts are from the same supplier?

Let \(X\) be a binomial random variable with \(p=0.1\) and \(n=10 .\) Calculate the following probabilities from the binomial probability mass function and also from the binomial table in Appendix A and compare results (a) \(P(X \leq 2)\) (b) \(P(X>8)\) (c) \(P(X=4)\) (d) \(P(5 \leq X \leq 7)\)

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Let \(X\) denote the number of bits received in error in a digital communication channel, and assume that \(X\) is a binomial random variable with \(p=0.001\). If 1000 bits are transmitted, determine the following: (a) \(P(X=1)\) (b) \(P(X \geq 1)\) (c) \(P(X \leq 2)\) (d) mean and variance of \(X\)

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